3. Equations involving Modulus

Hello.

This lesson will discuss the solution of equations involving modulus. I’ll talk about linear equations mainly. But the methods you’ll learn will be applicable across any type of equation, be it linear, quadratic, trigonometric, or any other type.

Example 1 Solve the equation |x| = 3.

Solution I’ll illustrate two different methods, both following from the definitions that we discussed previously.

Method 1

The definition of modulus tells me that .

So when x ≥ 0, |x| can be replaced by x, and therefore I’ve to solve the equation x = 3.

And, when x < 0, |x| equals –x, and hence I’ve to solve –x = 3.

The first equation gives the solution straight away, i.e. x = 3. The second one gives x = –3.

We therefore get two solutions x = 3 and –3.

You can substitute these solutions back in the original equation to verify: |3| = 3 and |–3| = 3.

Method 2

We’ll now use the second definition. Recall that |x| represents the distance of x from 0 on the number line.

We can therefore rephrase this problem as “Find a number whose distance from the origin is 3”.

Well, that’s easy enough. Let me draw the number line.

On this number line, we can clearly see that there are two numbers, 3 and – 3, that are at a distance of 3 from the origin.